\section{Construction elements of $K_2^T$ on a family of hyperelliptic curves}\label{section:element}
Let $C$ be the curve defined by the equation
\begin{equation}\label{eqn:curve}
y(y+\lambda \prod_{i=1}^g(a_i x+1))=x^{2g+2},
\end{equation}
where $\lambda, a_i \in \mathbb{Q}$. Consider the following elements in $K_2(\mathbb{Q}(C))$
\begin{equation}
M_i = \left\{ \frac{x^{2g+2}}{y^2}, a_i x + 1 \right\}, i=1,\ldots,g.
\end{equation}
We will prove in the following proposition these elements satisfy the tame symbol condition everywhere.

\begin{proposition}\label{prop:k2}
$M_i$ are elements of $K_2^T(C)$.
\end{proposition}
\begin{proof}
First we look at the points at infinity of the curve. Let $x=1/\tilde{x}$ and $y=\tilde{y}/\tilde{x}^{g+1}$. Then the equation (\ref{eqn:curve}) becomes
$$\tilde{y}(\tilde{y}+\lambda \tilde{x} \prod_{i=1}^g(\tilde{x}+a_i) )  = 1,$$
and the points at infinity become $(0, 1)$ and $(0,-1)$ which are denoted by $\infty$ and $\infty'$ respectively.

Let $P_{a_i},P_{a_i}'$ be the points $(-a_i^{-1},\pm(-a_i^{-1})^{g+1})$ and $O,O'$ be the points $(0,0), (0,-\lambda)$ respectively. Then we have
\begin{eqnarray*}
\divisor(a_i x+1) & = & (P_{a_i})+(P_{a_i}')-(\infty)-(\infty'). \\
\divisor \left(\frac{x^{2g+2}}{y^2}\right) &=& (2g+2)(O') - (2g+2)(O)
\end{eqnarray*}

Since $\frac{x^{2g+2}}{y^2}, a_i x+1$ only have zeros and poles at $P_{a_i},P_{a_i}', O, O', \infty$ and $\infty'$, the tame symbols of $M_{\beta_i}$ are trivial except at these points, for these points we have
\begin{eqnarray*}
T_{P_{a_i}}(M_i) & = & (-1)^0 \left.\frac{x^{2g+2}}{y^2}\right|_{P_{a_i}} = 1, \\
T_{P_{a_i}'}(M_i) & = & (-1)^0 \left.\frac{x^{2g+2}}{y^2}\right|_{P_{a_i}'} = 1, \\
T_{\infty}(M_i) & = & (-1)^0 \left.\left(\frac{x^{2g+2}}{y^2}\right)^{-1}\right|_{\infty} = \tilde{y}^2|_{\infty} = 1 , \\
T_{\infty'}(M_i) & = & (-1)^{0} \left.\left(\frac{x^{2g+2}}{y^2}\right)^{-1}\right|_{\infty'} = \tilde{y}^2|_{\infty'} = 1 , \\
T_{O}(M_i) & = & (-1)^0 \left. \frac{1}{(a_ix+1)^{-(2g+2)}} \right|_{O} =1,\\
T_{O'}(M_i) & = & (-1)^0 \left. \frac{1}{(a_ix+1)^{2g+2}}  \right|_{O'} = 1,
\end{eqnarray*}
which finishes the proof.
\end{proof}


\begin{theorem}\label{thm:main}
Let $C$ be defined by (\ref{eqn:curve}) where $\lambda$ and $a_1,\ldots,a_g$ be integers.
Denote the class of $M_i$ in $K_2^T(C)/\torsion$ also by $M_i$. Then $M_i \in K_2(C;\mathbb{Z})$.
\end{theorem}
\begin{proof}
Denote the arithmetic surface defined by (\ref{eqn:curve}) as $\mathcal{C}'$.
Let $\mathcal{C}$ be a regular proper model of $\mathcal{C}'$ and $\mathcal{C}_p$ be the fiber of $\mathcal{C}$ above $p$. To check integrality, we only need to show that for every irreducible component $\mathcal{D}$ of $\mathcal{C}_p$, $T_{\mathcal{D}}(M) = 1$. Since $\mathcal{C}$ is obtained from $\mathcal{C}'$ by consecutively blowing up singularities, every such $\mathcal{D}$ maps onto an irreducible component $D$ of the fiber above $p$ of $\mathcal{C}'$ or a singular point. We only need to examine the functions $f=\frac{x^{2g+2}}{y^2}$ and $g=a_ix+1$ at $D$ and singular points.

On any irreducible component $D$ of the fiber above $p$ of $\mathcal{C}'$, the functions $f$ and $g$ are rational functions which do not vanish or have a pole on it since it is obvious that $x,y$ does not vanish identically on $D$. So if $\mathcal{D}$ maps to an irreducible component, $v_{\mathcal{D}}(f) = 0, v_{\mathcal{D}}(g) = 0$, $T_{\mathcal{D}}(M) = 1$.

On the fiber above $p$ of $\mathcal{C}'$, I claim if $f$ or $g$ has a zero or pole then $g$ or $f$ equals 1. So if $\mathcal{D}$ maps to a point, the tame symbol at $\mathcal{D}$ vanishes. Actually at the point at infinity, $f=\tilde{y}^{-2}=1$ using the notation in \ref{prop:k2}, otherwise if $f$ has a zero or pole then $x=0$, $g=1$, if $g$ has a zero or pole then $y^2=x^{2g+2}$, $f=1$.

In sum, on any irreducible component $\mathcal{D}$ of the fiber above $p$ of $\mathcal{C}$, $T_\mathcal{D}(M)=1$.

\end{proof}
